Answer
$${\bf{r}}'\left( t \right) = \frac{1}{{1 + {t^2}}}{\bf{i}} + \left( {\cos t - t\sin t} \right){\bf{j}} - \frac{1}{{2\sqrt t }}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left( {{{\tan }^{ - 1}}t} \right){\bf{i}} + t\cos t{\bf{j}} - \sqrt t {\bf{k}} \cr
& {\text{Differentiate}} \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{\bf{r}}\left( t \right)} \right] \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left( {{{\tan }^{ - 1}}t} \right){\bf{i}} + t\cos t{\bf{j}} - \sqrt t {\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{{\tan }^{ - 1}}t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {t\cos t} \right]{\bf{j}} - \frac{d}{{dt}}\left[ {\sqrt t } \right]{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = \frac{1}{{1 + {t^2}}}{\bf{i}} + \left( {\cos t - t\sin t} \right){\bf{j}} - \frac{1}{{2\sqrt t }}{\bf{k}} \cr} $$
